使用PHP读取JSON POST [英] Reading JSON POST using PHP

问题描述

在发布此问题之前，我环顾了四周，因此我很抱歉在其他问题上发表，这只是我的第二个问题，如果无法正确设置此问题的格式，我们深表歉意.

I looked around a lot before posting this question so my apologies if it is on another post and this is only my second quesiton on here so apologies if I don't format this question correctly.

我创建了一个非常简单的Web服务，该服务需要获取发布值并返回JSON编码的数组.一切工作正常，直到被告知我需要使用内容类型为application/json的表单数据来发布.从那时起，我无法从Web服务返回任何值，这绝对与我如何过滤其发布值有关.

I have a really simple web service that I have created that needs to take post values and return a JSON encoded array. That all worked fine until I was told I would need to post the form data with a content-type of application/json. Since then I cannot return any values from the web service and it is definitely something to do with how I am filtering their post values.

基本上，在我的本地设置中，我创建了一个执行以下操作的测试页-

Basically in my local setup I have created a test page that does the following -

$curl = curl_init();
curl_setopt($curl, CURLOPT_CUSTOMREQUEST, "POST");                                                                     
curl_setopt($curl, CURLOPT_POSTFIELDS, $data);                                                                  
curl_setopt($curl, CURLOPT_RETURNTRANSFER, true);                                                                      
curl_setopt($curl, CURLOPT_HTTPHEADER, array(                                                                          
    'Content-Type: application/json',                                                                                
    'Content-Length: ' . strlen($data))                                                                       
);
curl_setopt($curl, CURLOPT_URL, 'http://webservice.local/');  // Set the url path we want to call
$result = curl_exec($curl);

//see the results
$json=json_decode($result,true);
curl_close($curl);
print_r($json);

在网络服务上，我有这个(我去除了一些功能)-

On the webservice I have this (I have stripped out some of the functions) -

<?php

header('Content-type: application/json');

/* connect to the db */
$link = mysql_connect('localhost','root','root') or die('Cannot connect to the DB');
mysql_select_db('webservice',$link) or die('Cannot select the DB');

if(isset($_POST['action']) && $_POST['action'] == 'login') {
    $statusCode = array('statusCode'=>1, 'statusDescription'=>'Login Process - Fail');
    $posts[] = array('status'=>$statusCode);
    header('Content-type: application/json');
    echo json_encode($posts);

    /* disconnect from the db */
}
@mysql_close($link);

?>

基本上我知道这是由于未设置$ _POST值，但是我找不到需要放置的内容而不是$ _POST.我试过了 json_decode($ _ POST)，file_get_contents("php://input")和许多其他方式，但是我在黑暗中开枪了.

Basically I know that it is due to the $_POST values not being set but I can't find what I need to put instead of the $_POST. I tried json_decode($_POST), file_get_contents("php://input") and a number of other ways but I was shooting in the dark a bit.

任何帮助将不胜感激.

谢谢，史蒂夫

感谢Michael的帮助，这是一个明确的进步，当我回应该帖子时，我现在至少有了一个回应....即使它为空

Thanks Michael for the help, that was a definite step forward I now have at least got a repsonse when I echo the post....even if it is null

更新的CURL-

  $curl = curl_init();
  curl_setopt($curl, CURLOPT_HTTPHEADER, array('Content-Type: application/json')); 
  curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST"); 
  curl_setopt($curl, CURLOPT_URL, 'http://webservice.local/');  
  curl_setopt($curl, CURLOPT_RETURNTRANSFER, true);
  curl_setopt($curl, CURLOPT_POSTFIELDS, json_encode($data));

数据发布到页面上的更新的php-

updated php on the page that the data is posted to -

$inputJSON = file_get_contents('php://input');
$input= json_decode( $inputJSON, TRUE ); //convert JSON into array

print_r(json_encode($input));

正如我所说，至少我现在看到的回应是在返回空白页之前

As I say at least I see a response now wheras prior it was returning a blank page

推荐答案

您有空的$_POST.如果您的网络服务器想要查看json格式的数据，则需要读取原始输入，然后使用JSON解码对其进行解析.

You have empty $_POST. If your web-server wants see data in json-format you need to read the raw input and then parse it with JSON decode.

您需要类似的东西:

$json = file_get_contents('php://input');
$obj = json_decode($json);

另外，您使用错误的代码来测试JSON通讯...

Also you have wrong code for testing JSON-communication...

CURLOPT_POSTFIELDS告诉curl将您的参数编码为application/x-www-form-urlencoded.您需要在这里使用JSON字符串.

CURLOPT_POSTFIELDS tells curl to encode your parameters as application/x-www-form-urlencoded. You need JSON-string here.

更新

您用于测试页的php代码应如下所示:

Your php code for test page should be like that:

$data_string = json_encode($data);

$ch = curl_init('http://webservice.local/');
curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST");
curl_setopt($ch, CURLOPT_POSTFIELDS, $data_string);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_HTTPHEADER, array(
        'Content-Type: application/json',
        'Content-Length: ' . strlen($data_string))
);

$result = curl_exec($ch);
$result = json_decode($result);
var_dump($result);

此外，在您的Web服务页面上，您还应删除header('Content-type: application/json');行之一.它只能被调用一次.

Also on your web-service page you should remove one of the lines header('Content-type: application/json');. It must be called only once.

这篇关于使用PHP读取JSON POST的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！